Once you have $n^2 + 2n - 48 = 0$, think of two numbers that multiply to $-48$ and add up to $2$.

After factoring, use the fact that if a product is $0$, then at least one factor must be $0$. This will give you two possible values for $n$; decide which one fits the condition in the question.

In Desmos, type y = x^2 + 2x - 48. This represents the equation after moving all terms to one side.

Look for the points where the graph crosses the x-axis (where $y = 0$). Note both x-values; these are the solutions to the equation, and the positive x-value is the one the question is asking for.

Start with the equation

To solve a quadratic equation, first set it equal to $0$ by subtracting $48$ from both sides:

We want to factor $n^2 + 2n - 48$ into $(n + a)(n + b)$.

Look for two numbers that:

• multiply to $-48$ (the constant term), and
• add to $2$ (the coefficient of $n$).

Those two numbers are $8$ and $-6$, so the factored form is

If a product of two factors is zero, then at least one factor must be zero. So set each factor equal to zero:

Solving these gives two values of $n$, one negative and one positive. The problem asks for the positive value of $n$, which is $6$, so the correct answer choice is C.